Question

BRAINLIEST PLEASEEE HELPLP 5. In a lab experiment, 2.5 grams of sodium bicarbonate is heated and decomposed into
sodium carbonate, carbon dioxide, and water vapor when heated. The actual yield of
sodium carbonate produced in the experiment is 2.04 grams. The theoretical yield of
each product is recorded in the data table below.
Using this data, determine the percent yield for sodium carbonate?
(Round Your Answer to the Nearest Whole Number)

Answer 1

Answer:

Explanation:

Sodium bicarbonate,

NaHCO

3

, will decompose to form sodium carbonate,

Na

2

CO

3

, water, and carbon dioxide,

CO

2

2

NaHCO

3(s]

→

Na

2

CO

3(s]

+

CO

2(g]

+

H

2

O

(g]

Notice that you have a

2

:

1

mole ratio between sodium bicarbonate and sodium carbonate. This means that the reaction will produce half as many moles of the latter than whatever number of moles of the former underwent decomposition.

Use sodium carbonate's molar amss to determine how many moles you'd get in that sample

0.685

g

⋅

1 mole NaHCO

3

84.007

g

=

0.008154 moles NaHCO

3

Now, if the reaction were to have a

100

%

yield, it would produce

0.008154

moles NaHCO

3

⋅

1 mole Na

2

CO

3

2

moles NaHCO

3

=

0.004077 moles Na

2

CO

3

Use the molar mass of sodium carbonate to determine how many grams would contain this many moles

0.004077

moles

⋅

105.99 g

1

mole

=

0.4321 g Na

2

CO

3Sodium bicarbonate,

NaHCO

3

, will decompose to form sodium carbonate,

Na

2

CO

3

, water, and carbon dioxide,

CO

2

2

NaHCO

3(s]

→

Na

2

CO

3(s]

+

CO

2(g]

+

H

2

O

(g]

Notice that you have a

2

:

1

mole ratio between sodium bicarbonate and sodium carbonate. This means that the reaction will produce half as many moles of the latter than whatever number of moles of the former underwent decomposition.

Use sodium carbonate's molar amss to determine how many moles you'd get in that sample

0.685

g

⋅

1 mole NaHCO

3

84.007

g

=

0.008154 moles NaHCO

3

Now, if the reaction were to have a

100

%

yield, it would produce

0.008154

moles NaHCO

3

⋅

1 mole Na

2

CO

3

2

moles NaHCO

3

=

0.004077 moles Na

2

CO

3

Use the molar mass of sodium carbonate to determine how many grams would contain this many moles

0.004077

moles

⋅

105.99 g

1

mole

=

0.4321 g Na

2

CO

3

Answer 2

By using the given data, the percent yield for sodium carbonate (Na₂CO₃) is equal to 127.

How to calculate percent yield?

Percent yield of any data can be calculated as:

% yield = (Actual value / Theoretical value) × 100

In the question actual yield of sodium carbonate is given, which is equal to 2.04 grams. And in the table theoretical yield of sodium carbonate also given, which is equal to 1.60 grams.

Now putting these value in the above equation, we get:

% yield = (2.04 / 1.60) × 100 = 127

Hence, percent yield of sodium carbonate (Na₂CO₃) is 127.

To learn more about percent yield, visit the below link:

https://brainly.com/question/11963853